By the AM-GM inequality, we have
$$\begin{aligned} \frac{bc}{a} + \frac{ca}{b} &= c\left(\frac{b}{a} + \frac{a}{b}\right) \ge 2c, \\ \frac{ca}{b} + \frac{ab}{c} &= a\left(\frac{c}{b} + \frac{b}{c}\right) \ge 2a, \\ \frac{ab}{c} + \frac{bc}{a} &= b\left(\frac{a}{c} + \frac{c}{a}\right) \ge 2b. \end{aligned}$$
Adding the three inequalities, we get
$$\frac{bc}{a} + \frac{ca}{b} + \frac{ab}{c} \ge a + b + c. \quad (1)$$
In addition, we have
$$a + b + c = \sqrt{(a^2 + b^2 + c^2) + 2(ab + bc + ca)} \ge \sqrt{3(ab + bc + ca)} \ge \sqrt{3}. \quad (2)$$
Combining (1) and (2), we obtain $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c} \ge \sqrt{3}$. This yields $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \ge \frac{\sqrt{3}}{abc}$.
Equality holds when $a=b=c=\frac{1}{\sqrt{3}}$.